Circular Motion, Explained
After reading this you will know how to turn a radius and any one measure of how fast something goes round (speed, period, frequency or RPM) into its centripetal acceleration, the force holding it in the circle, and its angular speed.
What circular motion actually is
An object moving in a circle at constant speed is still accelerating. That sounds wrong until you separate two ideas. Speed is how fast you cover distance. Velocity is speed plus direction. Going round a bend, your direction changes every instant, so your velocity changes even if the speedometer never moves. A changing velocity is an acceleration, and that acceleration points toward the centre of the circle.
Here is the hook. Take a 1 kg ball on a string, radius 2 m, whirling at 5 m/s. The string must pull it inward with about 12.5 newtons the whole time, roughly the weight of a 1.3 kg bag of sugar hanging from your hand. Double the speed to 10 m/s and the pull jumps to 50 N, four times larger. That square relationship is why a fast, tight turn feels so violent.
When to use this, and when not to
This tool assumes uniform circular motion: constant radius, constant speed, motion in a flat circle. That covers a lot of real problems. A car on a level circular track, a satellite in a near-circular orbit, a rotor blade tip, a mass on a string held horizontal, a point on a spinning disc.
It does not cover cases where the speed along the path is changing. A ball swinging on a string in a vertical circle speeds up at the bottom and slows at the top, so it also has a tangential acceleration that this model ignores. For that you need the full acceleration, which combines the centripetal part with a tangential part at right angles to it. The numbers here still give you the centripetal component at any instant if you plug in the speed at that instant.
Centripetal means "seeking the centre". It is not a new kind of force. It is the name for whatever real force happens to point inward: tension in a string, friction on tyres, gravity on a satellite, the normal force from a wall. If that inward force vanishes, the object flies off along a straight tangent, not outward.
The formulas and where they come from
Start with angular speed, written \omega (omega). It measures how many radians the object sweeps per second. One full turn is 2\pi radians, so if the period (time for one turn) is T:
Here f = 1/T is the frequency in hertz (turns per second). RPM is just frequency in different units: divide RPM by 60 to get hertz.
The linear speed v is how fast the object moves along the circle. In one period it travels the circumference 2\pi r, so:
Now the centripetal acceleration. Over a tiny time \Delta t the velocity vector rotates by the small angle \Delta\theta = \omega \, \Delta t. The change in the velocity vector has length v \, \Delta\theta, and it points inward. Dividing by \Delta t gives the acceleration:
All three forms are the same number, just written with whichever quantity you happen to know. The symbol a is the centripetal acceleration in m/s², v is speed in m/s, r is radius in metres, and \omega is angular speed in rad/s.
Finally, Newton's second law turns acceleration into force. Multiply by the mass m:
That F is the net inward force required, in newtons. It is not an extra force added to the diagram; it is the sum of the real forces resolved toward the centre.
A worked example with the demo numbers
A 1 kg mass on a 2 m string at 5 m/s
These are the tool's default values: radius 2 m, mass 1 kg, and a linear speed of 5 m/s. Press Calculate to reproduce every number below.
- Centripetal acceleration: a = v^2/r = 5^2/2 = 25/2 = 12.5 m/s².
- Centripetal force: F = ma = 1 \times 12.5 = 12.5 N.
- Angular speed: \omega = v/r = 5/2 = 2.5 rad/s.
- Period: T = 2\pi/\omega = 2\pi/2.5 \approx 2.513 s per revolution.
- Frequency: f = 1/T \approx 0.3979 Hz, which is about 0.3979 \times 60 \approx 23.87 RPM.
So the string tension is 12.5 N, the mass sweeps 2.5 radians every second, and it completes a full circle roughly every 2.5 seconds. Compare that 12.5 N to the ball's weight, mg \approx 9.81 N. The tension needed to keep it circling exceeds its own weight, which is why the string feels taut.
How the force grows with speed
The single most important feature of these formulas is the square. Force rises with the square of speed but only linearly with mass and inversely with radius. The table shows the required force on the same 1 kg mass at radius 2 m as speed climbs.
| Speed (m/s) | a = v²/r (m/s²) | F = mv²/r (N) |
|---|---|---|
| 2.5 | 3.125 | 3.125 |
| 5 | 12.5 | 12.5 |
| 10 | 50 | 50 |
| 15 | 112.5 | 112.5 |
| 20 | 200 | 200 |
Halving the speed from 5 to 2.5 m/s cuts the force to a quarter, not a half. Doubling it from 5 to 10 quadruples the force. This is the physics behind highway curve speed limits: the friction available from the tyres is roughly fixed, so the safe speed depends on the curve radius, and a small overspeed demands a large jump in grip.
See the trade-off yourself
Speed and radius pull in opposite directions. A wider circle at the same speed needs less force. Use the widget to feel how the two combine.
Reading and interpreting the results
The tool reports several quantities that describe the same motion. Keep their meanings straight.
- Centripetal acceleration
- The inward acceleration in m/s². It measures how sharply the velocity is turning, not how fast the object moves. A satellite in low Earth orbit has about 8.7 m/s² inward, close to gravity at the surface, yet it never gets closer to the ground.
- Centripetal force
- The net inward force in newtons needed to produce that acceleration on this mass. If the available force (grip, tension, gravity) is smaller, the object cannot stay on this circle and follows a wider path or flies off.
- Angular speed
- Radians per second, \omega = v/r. Two objects on the same rigid turntable share the same \omega but the outer one has a higher linear speed.
- Period and frequency
- Seconds per turn and turns per second. They are reciprocals: f = 1/T.
Common mistakes
There is no outward "centrifugal force" acting on the object in an inertial frame. The passenger thrown against a car door is not pushed outward; the door pushes them inward while their body tries to keep going straight. Adding a fictitious outward force to a free-body diagram double counts and gives wrong answers.
A second trap is mixing up angular and linear speed. If a problem gives you RPM, convert to \omega in rad/s before using a = \omega^2 r, or convert to v in m/s before using a = v^2/r. Do not put RPM straight into either formula. To go from RPM to rad/s, multiply by 2\pi/60 \approx 0.1047. So 100 RPM is about 10.47 rad/s.
A third is forgetting that radius is measured to the axis, not the diameter. If a wheel is 0.7 m across, its rim radius is 0.35 m. Using 0.7 by mistake doubles your radius and halves the acceleration you calculate.
A fourth: units of speed. A speed quoted as 90 km/h is 25 m/s. Feeding 90 into v^2/r overstates the acceleration by a factor of about 13.
Related tools
Circular motion sits inside the wider mechanics toolkit. To go from the centripetal force to whether a car actually holds a curve, compare it against the grip from the Friction & Inclined Plane Calculator. The step from force to acceleration is Newton's second law, handled directly by the Force Calculator. For motion in a straight line before and after the bend, use the Kinematics Calculator. If your object also spins about its own axis or is turned by an off-centre force, the Torque Calculator is the right one. And to check the energy in the moving mass, the Kinetic & Potential Energy Calculator gives \frac{1}{2}mv^2, which for the demo ball is 0.5 \times 1 \times 25 = 12.5 J.
Frequently asked questions
Does circular motion at constant speed do any work?
No. Work is force times distance along the direction of the force. The centripetal force points inward, while the motion is always tangential, at 90 degrees to it. Since \cos 90^\circ = 0, the work is zero and the kinetic energy stays constant. That is consistent with the speed never changing.
Why does the acceleration point to the centre if the object never reaches the centre?
Acceleration changes velocity, and here it changes only the direction of velocity, not its size. The inward pull constantly bends the straight-line motion into a curve. It is exactly enough to keep the object at the same radius, not to pull it in.
How do I convert RPM into the speed the tool uses?
First get angular speed: \omega = \text{RPM} \times 2\pi/60. Then linear speed is v = \omega r. For 60 RPM at radius 2 m: \omega = 60 \times 0.1047 \approx 6.283 rad/s and v = 6.283 \times 2 \approx 12.57 m/s.
What is the difference between angular speed and frequency?
Both describe how fast something turns, in different units. Frequency counts full turns per second, angular speed counts radians per second, and one turn is 2\pi radians. So \omega = 2\pi f. At 0.3979 Hz the angular speed is 2\pi \times 0.3979 \approx 2.5 rad/s.
Why does a tighter turn feel stronger at the same speed?
Because force is mv^2/r, and a smaller r means a larger force for the same speed and mass. Halving the radius doubles the required force. That is why a sharp exit ramp taken at highway speed pushes you far harder than a gentle motorway curve.
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